$\mathbf {Question:}$

Define a function $f:\Bbb R^2 \to \Bbb R$ by

$f(x,y)=$ $(x/|y|)\sqrt {x^2+y^2}$ if $y\not = 0$

$f(x,y)=0$ if $y=0$

$\mathbf{a)}$ prove that the function $f$ is not continuous at the point $(0,0)$


I need to find a sequence $\{u_k\}_{k\in \Bbb N}$ converges to $(0,0)$ such that $\{f(u_k)\}$ does not converge to $f(u)$

But I could not find such a sequence. Please can someone give me a hint about the sequence?

$\mathbf{b)}$ The function f has directional derivatives in all directions at the point $(0,0)$

$\mathbf{answer-b:}$ I need to prove that $\frac{\partial f}{\partial p} (0,0)=\,lim_{t\to 0}\frac{f((0,0)+tp)-f(0,0)}{t}$ exists.

Is this right? If this is true, please show me how to prove its existence?

$\mathbf{c)}$ prove that if $c$ is any number, then there is a vector $p$ of norm $1$ such that $\frac{\partial f}{\partial p} (0,0)=c$

$\mathbf{answer-c:}$ I could not produce any idea to solve the part.

Please can someone show and explain me my questions step by step? I am just starting real analysis and on my own. So I am confused so much :( thank you for helping :)



a) Take $(u_k)=(\frac{1}{k},\frac{1}{k^3})$.

b) If $p=(p_1,p_2)$ such that $p_2\neq 0$ $$\frac{\partial f}{\partial p} (0,0)=\lim_{t\to 0}\frac{f((0,0)+tp)-f(0,0)}{t}=\lim_{t\to 0}\frac{f((tp_1,tp_2))}{t}=\lim_{t\to 0}\frac{tp_1\sqrt{(tp_1)^2+(tp_2)^2}}{t|tp_2|}=\frac{p_1\sqrt{(p_1)^2+(p_2)^2}}{|p_2|}$$

and if $p=(p_1,0)$ then it's clear that $\frac{\partial f}{\partial p} (0,0)=0$

c) If $c=0$ take $p=(1,0)$ and if $c\neq 0$ then $$\frac{\partial f}{\partial p} (0,0)=\frac{p_1}{|p_2|}=c$$ and then $p_1=c|p_2|$ and since $p_1^2+p_2^2=c^2p_2^2+p_2^2=1$ then $$p=\left(\frac{c}{\sqrt{1+c^2}},\frac{1}{\sqrt{1+c^2}}\right)$$

  • $\begingroup$ $(u_k)\to 0$ as $k\to \infty$ okay! But $\{f(u_k)\} \to 0$ so f is continuous at (0,0) but I dont want f to be continuous here $\endgroup$ – B11b Jul 7 '13 at 1:08
  • 1
    $\begingroup$ $f(1/k,1/k)=(1/k)/(1/k\sqrt{2/k^2})=k^2/k\sqrt{2}\to\infty$ $\endgroup$ – user63181 Jul 7 '13 at 1:13
  • $\begingroup$ Aaow! I took the limit false:( sorry:( by the way, can you help me for part b and c? Please.. $\endgroup$ – B11b Jul 7 '13 at 1:15
  • 1
    $\begingroup$ Ok so I'll edit my answer. $\endgroup$ – user63181 Jul 7 '13 at 1:29
  • 1
    $\begingroup$ Enjoy with the answer and good luck. $\endgroup$ – user63181 Jul 7 '13 at 1:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.